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Jan 14, 2015 at 18:37 comment added Eric Wofsey More generally, both $K_1$ and $K_2$ must have the property that the Hurewicz map $\pi_2\to H_2$ vanishes, because any inclusion of 2-dimensional complexes is injective on $H_2$.
Jan 14, 2015 at 15:21 comment added Samarkand I have got almost the same answer on MSE (it was removed), of course, as Andre noticed it cannot work.
Jan 14, 2015 at 10:17 comment added André Henriques Indeed, this cannot work: there is no embeding from $RP^2\cup C$ into any 2-dimensional CW complex $Y$ such that the induced map $\pi_2(RP^2\cup C)\to\pi_2(Y)$ is zero. Easier example: there is no embeding from $S^2$ into any 2-dimensional CW complex $Y$ such that the induced map $\pi_2(S^2)\to\pi_2(Y)$ is zero. The reason is that any such embedding $S^2\to Y$ admits a retract $Y\to S^2$ up to homotopy.
Jan 14, 2015 at 8:56 comment added Grigory M Are you sure $K_2\to K_3$ induces a zero map on $\pi_2$? Why?
Jan 14, 2015 at 8:01 history answered user43326 CC BY-SA 3.0